Question: There are four points that are $5$ units from the line $y=13$ and $13$ units from the point $(7,13)$.  What is the sum of the $x$- and $y$-coordinates of all four of these points?
Let $(x,y)$ be one such point. Since $(x,y)$ is $5$ units from the line $y=13$, it must be 5 units above the line or 5 units below it.  This means that the $y$-coordinate is 8 or 18.  By the distance formula, since $(x,y)$ is 13 units from $(7,13)$, we have $$\sqrt{(x-7)^2+(y-13)^2}=13.$$Since $y$ is either 5 more or 5 less than 13, we know that $(y-13)^2=25$. So we substitute:

\begin{align*}
\sqrt{(x-7)^2+25}&=13\\
\Rightarrow\qquad (x-7)^2+25&=13^2\\
\Rightarrow\qquad (x-7)^2&=144\\
\Rightarrow\qquad x-7&=\pm 12.\\
\end{align*}So we have $x-7=12$ or $x-7=-12$, giving $x=19$ or $x=-5$.

Putting it all together, we have $y=8 \text{ or } 18$ and $x=-5\text{ or }19$, so our four possible points are $(-5,8),$ $(-5,18),$ $(19,8),$ and $(19,18).$  The sum of all of these coordinates is $\boxed{80}$.